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1 Scheme (Results) January 207 Pearson Edexcel International Advanced Subsidiary Level in Chemistry (WCH04) Paper 0 General Principles of Chemistry I Rates, Equilibria and Further Organic Chemistry (including synoptic assessment)

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes for employers. For further information, please visit our website at Our website subject pages hold useful resources, support material and live feeds from our subject advisors giving you access to a portal of information. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, for all kinds of people, wherever they are in the world. We ve been involved in education for over 50 years, and by working across 70 countries, in 00 languages, we have built an international reputation for our commitment to high standards and raising achievement through innovation in education. Find out more about how we can help you and your students at: January 70 Publications Code WCH04_70_MS* All the material in this publication is copyright Pearson Education Ltd 207

3 General ing Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should be used appropriately. All the marks on the mark scheme are designed to be awarded. Examiners should always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. schemes will indicate within the table where, and which strands of QWC, are being assessed. The strands are as follows: i) ensure that text is legible and that spelling, punctuation and grammar are accurate so that meaning is clear ii) select and use a form and style of writing appropriate to purpose and to complex subject matter iii) organise information clearly and coherently, using specialist vocabulary when appropriate

4 Using the Scheme Examiners should look for qualities to reward rather than faults to penalise. This does NOT mean giving credit for incorrect or inadequate answers, but it does mean allowing candidates to be rewarded for answers showing correct application of principles and knowledge. Examiners should therefore read carefully and consider every response: even if it is not what is expected it may be worthy of credit. The mark scheme gives examiners: an idea of the types of response expected how individual marks are to be awarded the total mark for each question examples of responses that should NOT receive credit. / means that the responses are alternatives and either answer should receive full credit. ( ) means that a phrase/word is not essential for the award of the mark, but helps the examiner to get the sense of the expected answer. Phrases/words in bold indicate that the meaning of the phrase or the actual word is essential to the answer. ecf/te/cq (error carried forward) means that a wrong answer given in an earlier part of a question is used correctly in answer to a later part of the same question. Candidates must make their meaning clear to the examiner to gain the mark. Make sure that the answer makes sense. Do not give credit for correct words/phrases which are put together in a meaningless manner. Answers must be in the correct context. Quality of Written Communication s which involve the writing of continuous prose will expect candidates to: write legibly, with accurate use of spelling, grammar and punctuation in order to make the meaning clear select and use a form and style of writing appropriate to purpose and to complex subject matter organise information clearly and coherently, using specialist vocabulary when appropriate. Full marks will be awarded if the candidate has demonstrated the above abilities. s where QWC is likely to be particularly important are indicated (QWC) in the mark scheme, but this does not preclude others.

5 Section A (multiple choice) Correct Answer A Unit should be (mol dm -3 s - ) divided by (mol dm -3 ) B Correct C Unit is not (mol dm -3 s - ) divided by (mol dm -3 ) D Unit is not (mol dm -3 s - ) divided by (mol dm -3 ) Correct Answer 2 A Rate decreases by factor of 4 when [NO] is halved and increases by factor of 2 when [Br 2 ] is doubled so overall decreases by factor of 2/ is halved B Rate is not doubled C Correct D Rate is not quartered Correct Answer 3 A k is not directly proportional to temperature B k does not decrease as temperature increases C Correct D k increases exponentially, not as shown Correct Answer 4 A The temperature drops so it is true that ΔH is positive B Correct C A gas is formed so it is true that ΔS system is positive D The reaction is spontaneous so it is true that ΔS total is positive Correct Answer 5 A The entropy of the system increases when more gas molecules form B The entropy of the system increases when a gas forms from a solid C Correct D The entropy of the system increases when solid turns to liquid

6 Correct Answer 6 A The enthalpy change for the equation shown is equivalent to providing the energy to form gaseous sodium and chloride ions (- Lattice energy) and then hydrating the ions (+ hydration energy) so sign of Lattice energy is incorrect B Sign of enthalpy change of hydration is incorrect C Sign of enthalpy change of hydration is incorrect D Correct Correct Answer 7 A The level of solubility is not the cause of the enthalpy change B The statement is true but does not explain the enthalpy change C The enthalpy change of hydration does not depend on the lattice energy D Correct Correct Answer 8 A The pressure of solids should not be included B The pressure of solids should not be included C Correct D The expression is upside down Correct Answer 9 A Correct B On warming more acid will dissociate so the ph will drop C On warming more acid will dissociate so [HCOOH] will decrease D On warming more acid will dissociate forming more methanoate ions Correct Answer 0 A The more concentrated NaOH will have a higher ph B Correct C Ammonia is a weaker base than NaOH so ph will be lower D Ammonia is a weaker base than NaOH so ph will be lower

7 Correct Answer A Correct B Weak acid/ strong base needs an indicator with a higher ph range C Weak acid/ weak base would not show a sharp change at ph 3.8 to 5.4 D Not an acid/ base titration Correct Answer 2 A Nitric acid is a proton acceptor here B - The HSO 4 ion is a proton acceptor here C These are both proton acceptors D Correct Correct Answer 3 A S N 2 means bi-molecular, not two step B Correct C A racemic mixture would form via a planar intermediate in S N, not in S N 2 D A transition state, not a planar intermediate, forms in S N 2 Correct Answer 4 A Ammonium ethanoate would form B Correct C The product is a cyanohydrin not ethanamide D Ethanamide would not form Correct Answer 5 A The acid needed is propanoic acid and the alcohol is 3- methylbutan-2-ol B The alcohol needed is 3-methylbutan-2-ol C The acid needed is propanoic acid D Correct

8 Correct Answer 6 A Propanone cannot be oxidised to an acid B Reduction of propanal would form an alcohol C Correct D The acid produced would be methanoic Correct Answer 7 A Correct B Both compounds contain C-C and C-H bonds only C Both compounds contain C-C,C-H, C-O and O-H bonds only D Both compounds contain C-C, C-H, C-O and C=O bonds only Correct Answer 8 A It is carried out at temperatures where samples have been vaporised B It cannot be used if the samples have decomposed C It a cannot be used if the samples cannot be vaporised D Correct Correct Answer 9 A C 2 H 2 Cl would have mass 6 with these isotopes B C 2 H 2 Cl would have mass 65 with these isotopes C C 2 H 2 Cl would have mass 65 with these isotopes D Correct Correct Answer 20 A Correct B No peak at cm - for a ketone C No peak at cm - for an alcohol D Alkane would not have a peak at 750 cm -

9 Section B 2a(i) (Concentration of) NaOH / OH - remains (almost) constant NaOH / OH - is in excess, so it does not limit rate Only the concentration of CV + changes significantly change in rate is dependent only on the change in CV + IGNE references to excess / increasing reliability / ensuring rate is suitable 2a(ii) Colorimetry / (use of) colorimeter Spectrophotometry Measurement of light absorbed Recognisable but incorrect spelling Calorimetry, ph measurement, conductivity, sampling, titration, quenching

10 2a(iii) 2 One half-life (shown on graph and measured correctly) as 7.5 ± 0.5 minutes () Second half-life also 7.5 ± 0.5 minutes () Half-lives do not need to be sequential. answers given on the graph Second halflife 5 minutes second half-life is the same as the first if a correct value for the first half-life has been given both half-lives are 7.5 ± 0.5 minutes scores (2)

11 2a(iv) First order () As half-life is constant As half-life is similar () If zero order or second order given then (0) marks Half-life stated to be constant but with different values in (iii) 2 2b(i) /T = 3.37 x 0-3 / and ln k = () /T = 3.36 x 0-3 ln k = Any answer not to 3 sf

12 2b(ii) 5 Graph: 3 marks First mark: axes correct with sensible scales i.e points/ line covering at least half the grid (3 squares horizontally and 3 squares vertically) and ln k values becoming more negative down the axis with negative signs shown. Horizontal axis at foot of graph () Vertical axis with ascending numbers more negative Second mark: Both axes labelled, with units on x axis: (/T)/0-3 K - (/T) x 0 3 /K x 0-3 etc with (/T) /K - (/T) x 0-3 /K - and just ln k and no units on y axis Missing brackets in expression for units () Third mark: points correctly plotted and best fit straight line drawn. () Allow if line covers points such that they do not show clearly but it is straight and gradient correct. IGNE extrapolation in either direction

13 Gradient: 2 marks. This may be shown on the graph Gradient in the range to (K) IGNE unit Negative sign (as long as a value has been calculated) () Value () Gradient calculated from data in table TE on incorrect plotting Value given as a fraction 2b(iii) E a = -(8.3 x = (+) 5204) = (+)52 kj mol - / (+) J mol - / 5.2 x 0 4 J mol - 2 MP Use of R x gradient () MP2 Value to 2sf and matching unit () TE from 2b(ii) kj /mol kj for kj mol - J for J mol - E a will be from +50 to +53 for gradients of to (Total for 2 = 4 marks)

14 22(a) Reagent: 2,4-dinitrophenylhydrazine Brady s reagent / 2,4-DNP(H) Formula: C 6 H 3 (NO 2 ) 2 NHNH 2 or with ring displayed () Dinitrile for dinitro 2 Result: yellow / orange / red AND precipitate / ppt / ppte / solid / crystals () combinations of these colours e.g. orangered, but NOT red-brown No TE on incorrect reagent 22(b) Reagent: iodine and sodium hydroxide 3 iodine in the presence of alkali iodine and hydroxide ions sodium chlorate(i) and potassium iodide () Result: (pale) yellow precipitate / solid / crystals medicinal / antiseptic smell (with P only) () Identity: triiodomethane / iodoform / CHI 3 () CH 3 I correct displayed formula IGNE additional organic product, even if incorrect Only allow TE if iodoform test or iodine given as reagent

15 22(c) 3-methylbutan-2-ol / 3-methyl-2-butanol 2-methylbutan-3-ol / 2-methyl-3-butanol Pentan--ol Pentan-2-ol IGNE formula 22d of peaks in low resolution nmr spectrum P Q 3 2 () 4 of H atoms producing peak with greatest area in low resolution nmr spectrum 6 6 () Splitting pattern of peak with greatest area in high resolution nmr spectrum Doublet () Duplet 2 (lines) Triplet () 3 (lines)

16 22e(i) Acceptable Answers Reject 4 MP Dipole on C=O () IGNE any dipole on attacking CN MP2 Arrow from lone pair on C of CN to carbon of C=O / to space between the CN and carbon of C=O and arrow from C=O bond to O or just beyond O IGNE Lone pairs on O () MP3 Correct intermediate including full negative charge on O () MP4 Arrow from oxygen to H and from H CN bond to C of CN Arrow from oxygen to H + Arrow from (anywhere on) oxygen to H of H 2 O and from H OH bond to OH IGNE Lone pairs on HCN IGNE missing / incorrect CN as other product () H + CN - C N may be written as CN

17 22e(ii) any named strong acid e.g. HCl / H 2 SO 4 Or Named weak acid e.g. ethanoic acid any named strong alkali /NaOH /KOH /OH - followed by an acid alkali and acid added at the same time IGNE water (eg HCl/H 2 O) IGNE reference to dilute / concentrated IGNE just dilute acid / H + / H 3 O + 22e(iii) 2 Displayed COO linkage between units () Rest of structure including extension bonds C 2 H 5 for CH 2 CH 3 COO at one end and no O at the other () IGNE Square brackets and subscript n Bond to CH 3 of the ethyl group Extra O at end (Total for 22 = 7 marks)

18 23a 2-hydroxypropanoic acid Just 2-hydroxypropanoic 23b Acceptable Answers MP Organic product with one OH substituted by Cl CH 3 CHClCOOH CH 3 CH(OH)COCl Can be displayed. () MP2 Second OH substituted CH 3 CHClCOCl () MP3 POCl 3 and HCl as products in balanced equation () CH 3 CH(OH)COOH + 2PCl 5 CH 3 CHClCOCl + 2POCl 3 + 2HCl Reject 3 MP3 available for balanced equation with any one OH replaced by Cl CH 3 CH(OH)COOH + PCl 5 CH 3 CHClCOOH + POCl 3 + HCl CH 3 CH(OH)COOH + PCl 5 CH 3 CH(OH)COCl + POCl 3 + HCl PCl 3 O for POCl 3

19 23c(i) K a = [CH 3 CH(OH)COO - ][H + ] [CH 3 CH(OH)COOH] + symbol instead of multiply on top line HA and A - for lactic acid and lactate if a key given H 3 O + for H + Round brackets instead of square brackets K a = [H + ] 2 [CH 3 CH(OH)COOH] 23c(ii) Data on K a for ethanoic acid pk a for both acids must be given Lactic acid is stronger /ethanoic acid is weaker AND EITHER Ethanoic acid has a lower K a =.7 x 0-5 / lactic acid has a higher K a than.7 x 0-5 Ethanoic acid has pk a = 4.8 AND lactic acid has pk a = 3.86 IGNE comments on degree of dissociation of the acids

20 23c(iii) Correct final answer without working scores both calculation marks. 4 [H + ] 2 = 2.07 x0-5 [H + ] based on [acid] = [salt] (giving ph = 3.86) for both marks [H + ] = (0.50) (.38 x 0-4 ) / (2.07 x 0-5 ) / 4.55 x 0-3 () ph = -log[h + ] = from quadratic () TE on incorrectly evaluated [H + ] as long as final ph < 7 e.g final ph = 2.80, if Ka for ethanoic acid used scores mark for the calculation. Assumption [H + ] = [CH 3 CH(OH)COO - ] H + is only from acid / no H + from ionization of water () Assumption 2 Ionization of the (weak) acid is negligible/ very small/ insignificant [CH 3 CH(OH)COOH] initial -x = [CH 3 CH(OH)COOH] eqm i for initial Just ionisation is negligible without reference to a compound [CH 3 CH(OH)COOH] initial = [CH 3 CH(OH)COOH] eqm [CH 3 CH(OH)COOH] eqm = 0.50 (mol dm -3 ) [H + ] << [HA] ()

21 23c(iv) Correct final answer = 4 marks 4 NB Rounding [lactate] to 0.2 moles gives mass = (g), which also scores 4 marks Method [H + ] in buffer = x 0-4 () [CH 3 CH(OH)COO - ] = K a x [CH 3 CH(OH)COOH] [H + ] = (.38 x 0-4 ) x (0.50) x 0-4 Rearrangement of equation to find [lactate] () [lactate] = (mol dm -3 ) () Mass required = x 2 = = 23.2 (g) Ignore sf except sf TE on incorrectly calculated [lactate] () 6.8 (g) because this is 0.5 x 2 Method 2 pk = ph log[salt]/[acid] 3.86 = 4.00 log[salt]/[acid] () log[salt]/[acid] = 0.4 [salt]/[acid] =.38 [acid]/[salt] = 0.72 () [salt] = (.38 x 0.5) = 0.207(mol dm -3 ) () Mass required = x 2 = = 23.2 (g) Ignore sf except sf () If clearly not [lactate] calculated, but [lactic acid], [OH - ] or [H + ]

22 *23c(v) IGNE discussion of buffer reaction with lactic acid and hydroxide ions 3 (large) reservoir of lactate ions (to combine with hydrogen ions) reservoir of sodium lactate (large) reservoir of conjugate base /salt if lactate ions shown in equation () CH 3 CH(OH)COO - + H + CH 3 CH(OH)COOH () Equation with sodium lactate Reaction reversed showing lactic acid dissociation Ratio of undissociated lactic acid : lactate is relatively unchanged Ratio of undissociated acid : (conjugate) base / salt is relatively unchanged () (Total for 23 =7 marks)

23 24a(i) K c = [NO] 2 [Cl 2 ] [NOCl] 2 IGNE State symbols Partial pressures Round brackets in place of square brackets + symbol instead of multiply on top line *24a(ii) MARK CONSEQUENTIALLY ON EXPRESSION IN (i) 4 NOCl NO(g) Cl 2 Mol at start Mol at eqm.780 (0.220) 0.0 () Concs /mol dm (= mols at eqm 5) This may be shown as mols at eqm 5 in K c expression () K c = ((0.044) 2 x (0.022)) = 3.36 x 0-4 mol dm -3 (0.356) 2 Value IGNE sf except sf () Units () independently, consistent with Kc expression in (i) Correct final answer without working scores 4 marks

24 *24a(iii) K c is the same as 2 EITHER temperature is unchanged it is unaffected by change is to volume / pressure / concentration () More NO (and Cl 2 ) is formed because the quotient of the K c expression decreases to keep K c constant More NO (and Cl 2 ) forms because the pressure is reduced, so the reaction goes to the side with more (gas) moles More NO (and Cl 2 ) forms because the pressure is reduced, so the reaction goes to the right independently () 24b(i) ΔH f S o NO Cl 2 0 (65.0) Blank space or a dash instead of 0 All three values (2) Any two values ()

25 24b(ii) Final answer of ΔH = (+) 77(.0) kj mol - scores 2 2 First mark : ΔH = (2x90.2) (2x5.7) Hess cycle 2NOCl(g) (2x5.7) 2NO(g) + Cl 2 (g) (2x90.2) N 2 (g)+o 2 (g)+cl 2 (g) () ΔH = (+) 77(.0) (kj mol - ) () IGNE Units Max () TE for using a value other than 0 for Cl 2 24b(iii) ΔS surroundings = - ΔH/T ΔS = - ΔH/T as long as there is reference to surroundings subsequently () 2 (As ΔH is positive), when T increases, ΔS surroundings becomes less negative (so ΔS total becomes less negative) IGNE smaller and decreasing for less negative () No TE for MP2 if answer to (ii) is negative 24b(iv) ΔS (total) = R lnk () 2 IGNE K c / K p K increases as T increases because EITHER ΔS (total) increases (as T increases) Equilibrium moves to the right (as T increases) ()

26 24c(i) 2 nd mark dependent on st, for both methods. EITHER Answers discussing entropy change, not entropy 2 (Kinetic) energy of each particle is greater () substances for particles So more ways of arranging particles or quanta / more disorder/ more random movement (at higher T) () IGNE More collisions At the higher temperature the Maxwell- Boltzmann curve is more spread out () So there is greater randomness in the distribution of energies/ speeds ()

27 F 24c(ii) and 24c(iii): if mol - is written as mol -, only penalise once 24c(ii) J mol - K - scores 2 marks 2 ΔS sys = ( (23.2) -2(305.5)) () Magnitude, sign and units () No TE on incorrect expression +63 J mol - K - for mark due to using data at 298K 24c(iii) Method ΔS surr = - ΔH/T 3 or use of expression e.g x 000/800 () Value of ΔS surr with sign and unit (-66.5 J mol - K - / kj mol - K - ) Value of ΔS total with sign and unit ( = J mol - K - / kj mol - K - ) () ΔS total negative so not spontaneous TE on incorrect ΔS values in (ii) and (iii) If this gives a positive value for ΔS total, then spontaneous () Method 2 When ΔS total = 0, then T ΔS system = ΔH () T= ΔH/ ΔS system = 53200/40.7 = 307 K () At T < 307 K reaction is not spontaneous () Method 3 ΔG = x 40.7 / = x (40.7/000) () = J mol - / kj mol - () ΔG positive so reaction is not spontaneous () (Total for 24 = 22 marks)

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